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Q.

A ball weighing 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for 0.01 sec. The average force exerted by the surface on the ball is:

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a

1 N

b

10 N

c

100 N

d

0.1N

answer is B.

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Detailed Solution

Fav=ΔpΔt=10×10−3(5+5)0.01=10N

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